kaz1
et tu
shaon0 said:If you tried to prove this by using logs it is:
Assume x^0=1
LHS=In(x^0) RHS=In(1)
LHS=0In(x) RHS=In(1)
LHS=0 RHS=0
Therefore, LHS=RHS
Thus, x^0=1
:uhhuh:shaon0 said:wat? i didn't do that.
shaon0 said:If you tried to prove this by using logs it is:
Assume x^0=1
LHS=In(x^0) RHS=In(1)
LHS=0In(x) RHS=In(1)
LHS=0 RHS=0
Therefore, LHS=RHS
Thus, x^0=1
:uhhuh:shaon0 said:wat? i didn't do that.
In(1)=0kaz1 said::uhhuh:
wat?darkliight said:ell n ...
It's logarithm naturale. ln(1)=0.shaon0 said:wat?
sorry my bad..Slidey said:It's logarithm naturale. ln(1)=0.
You wrote In(1)=0. L and I look similar in some fonts, but it's probably a good idea not to make that mistake.